Newtons Interpolationspolynom
Author: Prof. Dr. Dörte Haftendorn

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Newtons Interpolationspolynom

Ha Mai 2011

Lösung

A = (1, 2)
B = (2, 1.5)
C: (3, 4)
D: (4, 5)
E = (5, 4.5)
F = (5.5, 2)
N0(x) = 1
N1(x) = (N0(x) * (x - x(A)))
N2(x) = (N1(x) * (x - x(B)))
N3(x) = (N2(x) * (x - x(C)))
N4(x) = (N3(x) * (x - x(D)))
N5(x) = (N4(x) * (x - x(E)))
N6(x) = (N5(x) * (x - x(F)))
c0: y(A)
h0(x) = (c0 * N0(x))
c1: (y(B) - h0(x(B))) / N1(x(B))
h1(x) = h0(x) + (c1 * N1(x))
c2: (y(C) - h1(x(C))) / N2(x(C))
h2(x) = h1(x) + (c2 * N2(x))
c3: (y(D) - h2(x(D))) / N3(x(D))
h3(x) = h2(x) + (c3 * N3(x))
c4: (y(E) - h3(x(E))) / N4(x(E))
h4(x) = h3(x) + (c4 * N4(x))
c5: (y(F) - h4(x(F))) / N5(x(F))
h5(x) = h4(x) + (c5 * N5(x))
n: h5
a = false
k: 1
f(x) = h1(x) + (k * N2(x))
g(x) = h2(x) + N3(x)
h(x) = h3(x) + N4(x)
lo: h4